CHAPTER The zero vector. This vector has the property that for every vector, v, v + 0 = 0+v = v and λv = 0 if λ is the real number, zero.

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1 This is page 1 Printer: Opaque this CHAPTER 1 MULTILINEAR ALGEBRA 1.1 Background We will list below some definitions and theorems that are part of the curriculum of a standard theory-based sophomore level course in linear algebra. (Such a course is a prerequisite for reading these notes.) A vector space is a set, V, the elements of which we will refer to as vectors. It is equipped with two vector space operations: Vector space addition. Given two vectors, v 1 and v 2, one can add them to get a third vector, v 1 + v 2. Scalar multiplication. Given a vector, v, and a real number, λ, one can multiply v by λ to get a vector, λv. These operations satisfy a number of standard rules: associativity, commutativity, distributive laws, etc. which we assume you re familiar with. (See exercise 1 below.) In addition we ll assume you re familiar with the following definitions and theorems. 1. The zero vector. This vector has the property that for every vector, v, v + 0 = 0+v = v and λv = 0 if λ is the real number, zero. 2. Linear independence. A collection of vectors, v i, i = 1,...,k, is linearly independent if the map (1.1.1) R k V, (c 1,...,c k ) c 1 v c k v k is The spanning property. A collection of vectors, v i, i = 1,...,k, spans V if the map (1.1.1) is onto. 4. The notion of basis. The vectors, v i, in items 2 and 3 are a basis of V if they span V and are linearly independent; in other words, if the map (1.1.1) is bijective. This means that every vector, v, can be written uniquely as a sum (1.1.2) v = c i v i.

2 2 Chapter 1. Multilinear algebra 5. The dimension of a vector space. If V possesses a basis, v i, i = 1,...,k, V is said to be finite dimensional, and k is, by definition, the dimension of V. (It is a theorem that this definition is legitimate: every basis has to have the same number of vectors.) In this chapter all the vector spaces we ll encounter will be finite dimensional. 6. A subset, U, of V is a subspace if it s vector space in its own right, i.e., for v, v 1 and v 2 in U and λ in R, λv and v 1 + v 2 are in U. 7. Let V and W be vector spaces. A map, A : V W is linear if, for v, v 1 and v 2 in V and λ R (1.1.3) and (1.1.4) A(λv) = λav A(v 1 + v 2 ) = Av 1 + Av The kernel of A. This is the set of vectors, v, in V which get mapped by A into the zero vector in W. By (1.1.3) and (1.1.4) this set is a subspace of V. We ll denote it by Ker A. 9. The image of A. By (1.1.3) and (1.1.4) the image of A, which we ll denote by Im A, is a subspace of W. The following is an important rule for keeping track of the dimensions of KerA and Im A. (1.1.5) dim V = dim Ker A + dim Im A. Example 1. The map (1.1.1) is a linear map. The v i s span V if its image is V and the v i s are linearly independent if its kernel is just the zero vector in R k. 10. Linear mappings and matrices. Let v 1,...,v n be a basis of V and w 1,...,w m a basis of W. Then by (1.1.2) Av j can be written uniquely as a sum, m (1.1.6) Av j = c i,j w i, c i,j R. i=1 The m n matrix of real numbers, [c i,j ], is the matrix associated with A. Conversely, given such an m n matrix, there is a unique linear map, A, with the property (1.1.6).

3 1.1 Background An inner product on a vector space is a map B : V V R having the three properties below. (a) For vectors, v, v 1, v 2 and w and λ R B(v 1 + v 2,w) = B(v 1,w) + B(v 2,w) and B(λv,w) = λb(v,w). (b) For vectors, v and w, B(v,w) = B(w,v). (c) For every vector, v B(v,v) 0. Moreover, if v 0, B(v,v) is positive. Notice that by property (b), property (a) is equivalent to B(w,λv) = λb(w,v) and B(w,v 1 + v 2 ) = B(w,v 1 ) + B(w,v 2 ). The items on the list above are just a few of the topics in linear algebra that we re assuming our readers are familiar with. We ve highlighted them because they re easy to state. However, understanding them requires a heavy dollop of that indefinable quality mathematical sophistication, a quality which will be in heavy demand in the next few sections of this chapter. We will also assume that our readers are familiar with a number of more low-brow linear algebra notions: matrix multiplication, row and column operations on matrices, transposes of matrices, determinants of n n matrices, inverses of matrices, Cramer s rule, recipes for solving systems of linear equations, etc. (See 1.1 and 1.2 of Munkres book for a quick review of this material.)

4 4 Chapter 1. Multilinear algebra Exercises. 1. Our basic example of a vector space in this course is R n equipped with the vector addition operation (a 1,...,a n ) + (b 1,...,b n ) = (a 1 + b 1,...,a n + b n ) and the scalar multiplication operation λ(a 1,...,a n ) = (λa 1,...,λa n ). Check that these operations satisfy the axioms below. (a) Commutativity: v + w = w + v. (b) Associativity: u + (v + w) = (u + v) + w. (c) For the zero vector, 0 = (0,...,0), v + 0 = 0 + v. (d) v + ( 1)v = 0. (e) 1v = v. (f) (g) Associative law for scalar multiplication: (ab)v = a(bv). Distributive law for scalar addition: (a + b)v = av + bv. (h) Distributive law for vector addition: a(v + w) = av + aw. 2. Check that the standard basis vectors of R n : e 1 = (1,0,...,0), e 2 = (0,1,0,...,0), etc. are a basis. 3. Check that the standard inner product on R n B((a 1,...,a n ),(b 1,...,b n )) = n a i b i i=1 is an inner product. 1.2 Quotient spaces and dual spaces In this section we will discuss a couple of items which are frequently, but not always, covered in linear algebra courses, but which we ll need for our treatment of multilinear algebra in

5 The quotient spaces of a vector space 1.2 Quotient spaces and dual spaces 5 Let V be a vector space and W a vector subspace of V. A W-coset is a set of the form v + W = {v + w, w W }. It is easy to check that if v 1 v 2 W, the cosets, v 1 + W and v 2 + W, coincide while if v 1 v 2 W, they are disjoint. Thus the W-cosets decompose V into a disjoint collection of subsets of V. We will denote this collection of sets by V/W. One defines a vector addition operation on V/W by defining the sum of two cosets, v 1 + W and v 2 + W to be the coset (1.2.1) v 1 + v 2 + W and one defines a scalar multiplication operation by defining the scalar multiple of v + W by λ to be the coset (1.2.2) λv + W. It is easy to see that these operations are well defined. For instance, suppose v 1 + W = v 1 + W and v 2 + W = v 2 + W. Then v 1 v 1 and v 2 v 2 are in W; so (v 1 + v 2 ) (v 1 + v 2 ) is in W and hence v 1 + v 2 + W = v 1 + v 2 + W. These operations make V/W into a vector space, and one calls this space the quotient space of V by W. We define a mapping (1.2.3) π : V V/W by setting π(v) = v + W. It s clear from (1.2.1) and (1.2.2) that π is a linear mapping, and that it maps V to V/W. Moreover, for every coset, v + W, π(v) = v + W; so the mapping, π, is onto. Also note that the zero vector in the vector space, V/W, is the zero coset, 0+W = W. Hence v is in the kernel of π if v +W = W, i.e., v W. In other words the kernel of π is W. In the definition above, V and W don t have to be finite dimensional, but if they are, then (1.2.4) dim V/W = dim V dim W. by (1.1.5). The following, which is easy to prove, we ll leave as an exercise.

6 6 Chapter 1. Multilinear algebra Proposition Let U be a vector space and A : V U a linear map. If W KerA there exists a unique linear map, A # : V/W U with property, A = A # π. The dual space of a vector space We ll denote by V the set of all linear functions, l : V R. If l 1 and l 2 are linear functions, their sum, l 1 + l 2, is linear, and if l is a linear function and λ is a real number, the function, λl, is linear. Hence V is a vector space. One calls this space the dual space of V. Suppose V is n-dimensional, and let e 1,...,e n be a basis of V. Then every vector, v V, can be written uniquely as a sum Let v = c 1 e c n e n c i R. (1.2.5) e i (v) = c i. If v = c 1 e c n e n and v = c 1 e c ne n then v + v = (c 1 + c 1 )e (c n + c n)e n, so e i (v + v ) = c i + c i = e i (v) + e i (v ). This shows that e i (v) is a linear function of v and hence e i V. Claim: e i, i = 1,...,n is a basis of V. Proof. First of all note that by (1.2.5) (1.2.6) e i (e j ) = { 1, i = j 0, i j. If l V let λ i = l(e i ) and let l = λ i e i. Then by (1.2.6) (1.2.7) l (e j ) = λ i e i (e j ) = λ j = l(e j ), i.e., l and l take identical values on the basis vectors, e j. Hence l = l. Suppose next that λ i e i = 0. Then by (1.2.6), λ j = ( λ i e i )(e j) = 0 for all j = 1,...,n. Hence the e j s are linearly independent.

7 1.2 Quotient spaces and dual spaces 7 Let V and W be vector spaces and A : V W, a linear map. Given l W the composition, l A, of A with the linear map, l : W R, is linear, and hence is an element of V. We will denote this element by A l, and we will denote by A : W V the map, l A l. It s clear from the definition that A (l 1 + l 2 ) = A l 1 + A l 2 and that A λl = λa l, i.e., that A is linear. Definition. A is the transpose of the mapping A. We will conclude this section by giving a matrix description of A. Let e 1,...,e n be a basis of V and f 1,...,f m a basis of W; let e 1,...,e n and f 1,...,f m be the dual bases of V and W. Suppose A is defined in terms of e 1,...,e n and f 1,...,f m by the m n matrix, [a i,j ], i.e., suppose Ae j = a i,j f i. Claim. A is defined, in terms of f 1,...,f m and e 1,...,e n by the transpose matrix, [a j,i ]. Proof. Let A f i = c j,i e j. Then A f i (e j) = k c k,i e k (e j) = c j,i by (1.2.6). On the other hand ( ) A fi (e j ) = fi (Ae j ) = fi ak,j f k = k a k,j f i (f k ) = a i,j so a i,j = c j,i.

8 8 Chapter 1. Multilinear algebra Exercises. 1. Let V be an n-dimensional vector space and W a k-dimensional subspace. Show that there exists a basis, e 1,...,e n of V with the property that e 1,...,e k is a basis of W. Hint: Induction on n k. To start the induction suppose that n k = 1. Let e 1,...,e n 1 be a basis of W and e n any vector in V W. 2. In exercise 1 show that the vectors f i = π(e k+i ), i = 1,...,n k are a basis of V/W. 3. In exercise 1 let U be the linear span of the vectors, e k+i, i = 1,...,n k. Show that the map U V/W, u π(u), is a vector space isomorphism, i.e., show that it maps U bijectively onto V/W. 4. Let U, V and W be vector spaces and let A : V W and B : U V be linear mappings. Show that (AB) = B A. 5. Let V = R 2 and let W be the x 1 -axis, i.e., the one-dimensional subspace {(x 1,0); x 1 R} of R 2. (a) Show that the W-cosets are the lines, x 2 = a, parallel to the x 1 -axis. (b) Show that the sum of the cosets, x 2 = a and x 2 = b is the coset x 2 = a + b. (c) Show that the scalar multiple of the coset, x 2 = c by the number, λ, is the coset, x 2 = λc. 6. (a) Let (V ) be the dual of the vector space, V. For every v V, let µ v : V R be the function, µ v (l) = l(v). Show that the µ v is a linear function on V, i.e., an element of (V ), and show that the map (1.2.8) µ : V (V ) v µ v is a linear map of V into (V ).

9 1.2 Quotient spaces and dual spaces 9 (b) Show that the map (1.2.8) is bijective. (Hint: dim(v ) = dimv = dimv, so by (1.1.5) it suffices to show that (1.2.8) is injective.) Conclude that there is a natural identification of V with (V ), i.e., that V and (V ) are two descriptions of the same object. 7. Let W be a vector subspace of V and let W = {l V, l(w) = 0 if w W }. Show that W is a subspace of V and that its dimension is equal to dimv dimw. (Hint: By exercise 1 we can choose a basis, e 1,...,e n of V such that e 1,...e k is a basis of W. Show that e k+1,...,e n is a basis of W.) W is called the annihilator of W in V. 8. Let V and V be vector spaces and A : V V a linear map. Show that if W is the kernel of A there exists a linear map, B : V/W V, with the property: A = B π, π being the map (1.2.3). In addition show that this linear map is injective. 9. Let W be a subspace of a finite-dimensional vector space, V. From the inclusion map, ι : W V, one gets a transpose map, ι : (V ) (W ) and, by composing this with (1.2.8), a map ι µ : V (W ). Show that this map is onto and that its kernel is W. Conclude from exercise 8 that there is a natural bijective linear map ν : V/W (W ) with the property ν π = ι µ. In other words, V/W and (W ) are two descriptions of the same object. (This shows that the quotient space operation and the dual space operation are closely related.) 10. Let V 1 and V 2 be vector spaces and A : V 1 V 2 a linear map. Verify that for the transpose map: A : V2 V 1 and KerA = (Im A) ImA = (Ker A).

10 10 Chapter 1. Multilinear algebra 11. (a) Let B : V V R be an inner product on V. For v V let l v : V R be the function: l v (w) = B(v,w). Show that l v is linear and show that the map (1.2.9) L : V V, v l v is a linear mapping. (b) Prove that this mapping is bijective. (Hint: Since dim V = dim V it suffices by (1.1.5) to show that its kernel is zero. Now note that if v 0 l v (v) = B(v,v) is a positive number.) Conclude that if V has an inner product one gets from it a natural identification of V with V. 12. Let V be an n-dimensional vector space and B : V V R an inner product on V. A basis, e 1,...,e n of V is orthonormal is { 1 i = j (1.2.10) B(e i,e j ) = 0 i j (a) Show that an orthonormal basis exists. Hint: By induction let e i, i = 1,...,k be vectors with the property (1.2.10) and let v be a vector which is not a linear combination of these vectors. Show that the vector w = v B(e i,v)e i is non-zero and is orthogonal to the e i s. Now let e k+1 = λw, where λ = B(w,w) 1 2. (b) Let e 1,...e n and e 1,... e n be two orthogonal bases of V and let (1.2.11) e j = a i,j e i. Show that (1.2.12) ai,j a i,k = { 1 j = k 0 j k (c) Let A be the matrix [a i,j ]. Show that (1.2.12) can be written more compactly as the matrix identity (1.2.13) AA t = I where I is the identity matrix.

11 1.2 Quotient spaces and dual spaces 11 (d) Let e 1,...,e n be an orthonormal basis of V and e 1,...,e n the dual basis of V. Show that the mapping (1.2.9) is the mapping, Le i = e i, i = 1,... n.

12 12 Chapter 1. Multilinear algebra 1.3 Tensors Let V be an n-dimensional vector space and let V k be the set of all k-tuples, (v 1,...,v k ), v i V. A function T : V k R is said to be linear in its i th variable if, when we fix vectors, v 1,...,v i 1, v i+1,...,v k, the map (1.3.1) v V T(v 1,...,v i 1,v,v i+1,...,v k ) is linear in V. If T is linear in its i th variable for i = 1,...,k it is said to be k-linear, or alternatively is said to be a k-tensor. We denote the set of all k-tensors by L k (V ). We will agree that 0-tensors are just the real numbers, that is L 0 (V ) = R. Let T 1 and T 2 be functions on V k. It is clear from (1.3.1) that if T 1 and T 2 are k-linear, so is T 1 + T 2. Similarly if T is k-linear and λ is a real number, λt is k-linear. Hence L k (V ) is a vector space. Note that for k = 1, k-linear just means linear, so L 1 (V ) = V. Let I = (i 1,...i k ) be a sequence of integers with 1 i r n, r = 1,...,k. We will call such a sequence a multi-index of length k. For instance the multi-indices of length 2 are the square arrays of pairs of integers (i,j), 1 i,j n and there are exactly n 2 of them. Exercise. Show that there are exactly n k multi-indices of length k. Now fix a basis, e 1,...,e n, of V and for T L k (V ) let (1.3.2) T I = T(e i1,...,e ik ) for every multi-index I of length k. Proposition The T I s determine T, i.e., if T and T are k-tensors and T I = T I for all I, then T = T.

13 1.3 Tensors 13 Proof. By induction on n. For n = 1 we proved this result in 1.1. Let s prove that if this assertion is true for n 1, it s true for n. For each e i let T i be the (k 1)-tensor Then for v = c 1 e 1 + c n e n (v 1,...,v n 1 ) T(v 1,...,v n 1,e i ). T(v 1,...,v n 1,v) = c i T i (v 1,...,v n 1 ), so the T i s determine T. Now apply induction. The tensor product operation If T 1 is a k-tensor and T 2 is an l-tensor, one can define a k+l-tensor, T 1 T 2, by setting (T 1 T 2 )(v 1,...,v k+l ) = T 1 (v 1,...,v k )T 2 (v k+1,...,v k+l ). This tensor is called the tensor product of T 1 and T 2. We note that if T 1 or T 2 is a 0-tensor, i.e., scalar, then tensor product with it is just scalar multiplication by it, that is a T = T a = at (a R, T L k (V )). Similarly, given a k-tensor, T 1, an l-tensor, T 2 and an m-tensor, T 3, one can define a (k + l + m)-tensor, T 1 T 2 T 3 by setting (1.3.3) T 1 T 2 T 3 (v 1,...,v k+l+m ) = T 1 (v 1,...,v k )T 2 (v k+1,...,v k+l )T 3 (v k+l+1,...,v k+l+m ). Alternatively, one can define (1.3.3) by defining it to be the tensor product of T 1 T 2 and T 3 or the tensor product of T 1 and T 2 T 3. It s easy to see that both these tensor products are identical with (1.3.3): (1.3.4) (T 1 T 2 ) T 3 = T 1 (T 2 T 3 ) = T 1 T 2 T 3. We leave for you to check that if λ is a real number (1.3.5) λ(t 1 T 2 ) = (λt 1 ) T 2 = T 1 (λt 2 ) and that the left and right distributive laws are valid: For k 1 = k 2, (1.3.6) (T 1 + T 2 ) T 3 = T 1 T 3 + T 2 T 3

14 14 Chapter 1. Multilinear algebra and for k 2 = k 3 (1.3.7) T 1 (T 2 + T 3 ) = T 1 T 2 + T 1 T 3. A particularly interesting tensor product is the following. For i = 1,...,k let l i V and let (1.3.8) T = l 1 l k. Thus, by definition, (1.3.9) T(v 1,...,v k ) = l 1 (v 1 )... l k (v k ). A tensor of the form (1.3.9) is called a decomposable k-tensor. These tensors, as we will see, play an important role in what follows. In particular, let e 1,...,e n be a basis of V and e 1,...,e n the dual basis of V. For every multi-index, I, of length k let e I = e i 1 e i k. Then if J is another multi-index of length k, { e 1, I = J (1.3.10) I(e j1,...,e jk ) = 0, I J by (1.2.6), (1.3.8) and (1.3.9). From (1.3.10) it s easy to conclude Theorem The e I s are a basis of Lk (V ). Proof. Given T L k (V ), let T = T I e I where the T I s are defined by (1.3.2). Then (1.3.11) T (e j1,...,e jk ) = T I e I (e j 1,...,e jk ) = T J by (1.3.10); however, by Proposition the T J s determine T, so T = T. This proves that the e I s are a spanning set of vectors for L k (V ). To prove they re a basis, suppose CI e I = 0 for constants, C I R. Then by (1.3.11) with T = 0, C J = 0, so the e I s are linearly independent. As we noted above there are exactly n k multi-indices of length k and hence n k basis vectors in the set, {e I }, so we ve proved Corollary. dim L k (V ) = n k.

15 1.3 Tensors 15 The pull-back operation Let V and W be finite dimensional vector spaces and let A : V W be a linear mapping. If T L k (W), we define to be the function A T : V k R (1.3.12) A T(v 1,...,v k ) = T(Av 1,...,Av k ). It s clear from the linearity of A that this function is linear in its i th variable for all i, and hence is k-tensor. We will call A T the pull-back of T by the map, A. Proposition The map (1.3.13) A : L k (W) L k (V ), T A T, is a linear mapping. We leave this as an exercise. We also leave as an exercise the identity (1.3.14) A (T 1 T 2 ) = A T 1 A T 2 for T 1 L k (W) and T 2 L m (W). Also, if U is a vector space and B : U V a linear mapping, we leave for you to check that (1.3.15) (AB) T = B (A T) for all T L k (W). Exercises. 1. Verify that there are exactly n k multi-indices of length k. 2. Prove Proposition Verify (1.3.14). 4. Verify (1.3.15).

16 16 Chapter 1. Multilinear algebra 5. Let A : V W be a linear map. Show that if l i, i = 1,...,k are elements of W A (l 1 l k ) = A l 1 A l k. Conclude that A maps decomposable k-tensors to decomposable k-tensors. 6. Let V be an n-dimensional vector space and l i, i = 1,2, elements of V. Show that l 1 l 2 = l 2 l 1 if and only if l 1 and l 2 are linearly dependent. (Hint: Show that if l 1 and l 2 are linearly independent there exist vectors, v i, i =,1,2 in V with property l i (v j ) = { 1, i = j 0, i j. Now compare (l 1 l 2 )(v 1,v 2 ) and (l 2 l 1 )(v 1,v 2 ).) Conclude that if dim V 2 the tensor product operation isn t commutative, i.e., it s usually not true that l 1 l 2 = l 2 l Let T be a k-tensor and v a vector. Define T v : V k 1 R to be the map (1.3.16) T v (v 1,...,v k 1 ) = T(v,v 1,...,v k 1 ). Show that T v is a (k 1)-tensor. 8. Show that if T 1 is an r-tensor and T 2 is an s-tensor, then if r > 0, (T 1 T 2 ) v = (T 1 ) v T Let A : V W be a linear map mapping v V to w W. Show that for T L k (W), A (T w ) = (A T) v.

17 1.4 Alternating k-tensors Alternating k-tensors We will discuss in this section a class of k-tensors which play an important role in multivariable calculus. In this discussion we will need some standard facts about the permutation group. For those of you who are already familiar with this object (and I suspect most of you are) you can regard the paragraph below as a chance to refamiliarize yourselves with these facts. Permutations Let k be the k-element set: {1,2,...,k}. A permutation of order k is a bijective map, σ : k k. Given two permutations, σ 1 and σ 2, their product, σ 1 σ 2, is the composition of σ 1 and σ 2, i.e., the map, i σ 1 (σ 2 (i)), and for every permutation, σ, one denotes by σ 1 the inverse permutation: σ(i) = j σ 1 (j) = i. Let S k be the set of all permutations of order k. One calls S k the permutation group of k or, alternatively, the symmetric group on k letters. Check: There are k! elements in S k. For every 1 i < j k, let τ = τ i,j be the permutation (1.4.1) τ(i) = j τ(j) = i τ(l) = l, l i,j. τ is called a transposition, and if j = i + 1, τ is called an elementary transposition. Theorem Every permutation can be written as a product of finite number of transpositions.

18 18 Chapter 1. Multilinear algebra Proof. Induction on k: k = 2 is obvious. The induction step: k 1 implies k : Given σ S k, σ(k) = i τ ik σ(k) = k. Thus τ ik σ is, in effect, a permutation of k 1. By induction, τ ikσ can be written as a product of transpositions, so σ = τ ik (τ ik σ) can be written as a product of transpositions. Theorem Every transposition can be written as a product of elementary transpositions. Proof. Let τ = τ ij, i < j. With i fixed, argue by induction on j. Note that for j > i + 1 Now apply induction to τ i,j 1. τ ij = τ j 1,j τ i,j 1 τ j 1,j. Corollary. Every permutation can be written as a product of elementary transpositions. The sign of a permutation Let x 1,...,x k be the coordinate functions on R k. For σ S k we define (1.4.2) ( 1) σ = i<j x σ(i) x σ(j) x i x j. Notice that the numerator and denominator in this expression are identical up to sign. Indeed, if p = σ(i) < σ(j) = q, the term, x p x q occurs once and just once in the numerator and one and just one in the denominator; and if q = σ(i) > σ(j) = p, the term, x p x q, occurs once and just once in the numerator and its negative, x q x p, once and just once in the numerator. Thus (1.4.3) ( 1) σ = ±1.

19 1.4 Alternating k-tensors 19 Claim: For σ,τ S k (1.4.4) ( 1) στ = ( 1) σ ( 1) τ. Proof. By definition, ( 1) στ = i<j x στ(i) x στ(j) x i x j. We write the right hand side as a product of (1.4.5) i<j x τ(i) x τ(j) x i x j = ( 1) τ and (1.4.6) x στ(i) x στ(j) x τ(i) x τ(j) i<j For i < j, let p = τ(i) and q = τ(j) when τ(i) < τ(j) and let p = τ(j) and q = τ(i) when τ(j) < τ(i). Then x στ(i) x στ(j) x τ(i) x τ(j) = x σ(p) x σ(q) x p x q (i.e., if τ(i) < τ(j), the numerator and denominator on the right equal the numerator and denominator on the left and, if τ(j) < τ(i) are negatives of the numerator and denominator on the left). Thus (1.4.6) becomes x σ(p) x σ(q) = ( 1) σ. x p x q p<q We ll leave for you to check that if τ is a transposition, ( 1) τ = 1 and to conclude from this: Proposition If σ is the product of an odd number of transpositions, ( 1) σ = 1 and if σ is the product of an even number of transpositions ( 1) σ = +1.

20 20 Chapter 1. Multilinear algebra Alternation Let V be an n-dimensional vector space and T L (v) a k-tensor. If σ S k, let T σ L (V ) be the k-tensor (1.4.7) T σ (v 1,...,v k ) = T(v σ 1 (1),...,v σ 1 (k)). Proposition If T = l 1 l k, l i V, then T σ = l σ(1) l σ(k). 2. The map, T L k (V ) T σ L k (V ) is a linear map. 3. T στ = (T τ ) σ. Proof. To prove 1, we note that by (1.4.7) (l 1 l k ) σ (v 1,...,v k ) = l 1 (v σ 1 (1)) l k (v σ 1 (k)). Setting σ 1 (i) = q, the i th term in this product is l σ(q) (v q ); so the product can be rewritten as l σ(1) (v 1 )... l σ(k) (v k ) or (l σ(1) l σ(k) )(v 1,...,v k ). The proof of 2 we ll leave as an exercise. Proof of 3: By item 2, it suffices to check 3 for decomposable tensors. However, by 1 (l 1 l k ) στ = l στ(1) l στ(k) = (l τ(1) l τ(k) ) σ = ((l 1 l) τ ) σ. Definition T L k (V ) is alternating if T σ = ( 1) σ T for all σ S k. We will denote by A k (V ) the set of all alternating k-tensors in L k (V ). By item 2 of Proposition this set is a vector subspace of L k (V ).

21 1.4 Alternating k-tensors 21 It is not easy to write down simple examples of alternating k- tensors; however, there is a method, called the alternation operation, for constructing such tensors: Given T L (V ) let (1.4.8) Alt T = τ S k ( 1) τ T τ. We claim Proposition For T L k (V ) and σ S k, 1. (Alt T) σ = ( 1) σ Alt T 2. if T A k (V ), AltT = k!t. 3. Alt T σ = (Alt T) σ 4. the map is linear. Alt : L k (V ) L k (V ), T Alt (T) Proof. To prove 1 we note that by Proposition (1.4.4): (Alt T) σ = ( 1) τ (T στ ) = ( 1) σ ( 1) στ T στ. But as τ runs over S k, στ runs over S k, and hence the right hand side is ( 1) σ Alt (T). Proof of 2. If T A k Alt T = ( 1) τ T τ = ( 1) τ ( 1) τ T = k!t. Proof of 3. Alt T σ = ( 1) τ T τσ = ( 1) σ ( 1) τσ T τσ = ( 1) σ Alt T = (Alt T) σ.

22 22 Chapter 1. Multilinear algebra Finally, item 4 is an easy corollary of item 2 of Proposition We will use this alternation operation to construct a basis for A k (V ). First, however, we require some notation: Let I = (i 1,...,i k ) be a multi-index of length k. Definition I is repeating if i r = i s for some r s. 2. I is strictly increasing if i 1 < i 2 < < i r. 3. For σ S k, I σ = (i σ(1),...,i σ(k) ). Remark: If I is non-repeating there is a unique σ S k so that I σ is strictly increasing. Let e 1,...,e n be a basis of V and let e I = e i 1 e i k and ψ I = Alt (e I). Proposition ψ I σ = ( 1) σ ψ I. 2. If I is repeating, ψ I = If I and J are strictly increasing, ψ I (e j1,...,e jk ) = { 1 I = J 0 I J. Proof. To prove 1 we note that (e I )σ = e Iσ; so Alt (e I σ) = Alt (e I) σ = ( 1) σ Alt (e I). Proof of 2: Suppose I = (i 1,...,i k ) with i r = i s for r s. Then if τ = τ ir,i s, e I = e I r so ψ I = ψ I r = ( 1) τ ψ I = ψ I.

23 1.4 Alternating k-tensors 23 Proof of 3: By definition But by (1.3.10) ψ I (e j1,...,e jk ) = ( 1) τ e I τ(e j 1,...,e jk ). (1.4.9) e I τ(e j 1,...,e jk ) = { 1 if I τ = J 0 if I τ J. Thus if I and J are strictly increasing, I τ is strictly increasing if and only if I τ = I, and (1.4.9) is non-zero if and only if I = J. Now let T be in A k. By Proposition 1.3.2, Since T = a J e J, a J R. k!t = Alt (T) T = 1 k! aj Alt (e J ) = b J ψ J. We can discard all repeating terms in this sum since they are zero; and for every non-repeating term, J, we can write J = I σ, where I is strictly increasing, and hence ψ J = ( 1) σ ψ I. Conclusion: We can write T as a sum (1.4.10) T = c I ψ I, with I s strictly increasing. Claim. The c I s are unique.

24 24 Chapter 1. Multilinear algebra Proof. For J strictly increasing (1.4.11) T(e j1,...,e jk ) = c I ψ I (e j1,...,e jk ) = c J. By (1.4.10) the ψ I s, I strictly increasing, are a spanning set of vectors for A k (V ), and by (1.4.11) they are linearly independent, so we ve proved Proposition The alternating tensors, ψ I, I strictly increasing, are a basis for A k (V ). Thus dim A k (V ) is equal to the number of strictly increasing multiindices, I, of length k. We leave for you as an exercise to show that this number is equal to ( ) n n! (1.4.12) = = n choose k k (n k)!k! if 1 k n. Hint: Show that every strictly increasing multi-index of length k determines a k element subset of {1,..., n} and vice-versa. Note also that if k > n every multi-index I = (i 1,...,i k ) of length k has to be repeating: i r = i s for some r s since the i p s lie on the interval 1 i n. Thus by Proposition ψ I = 0 for all multi-indices of length k > 0 and (1.4.13) A k = {0}. Exercises. 1. Show that there are exactly k! permutations of order k. Hint: Induction on k: Let σ S k, and let σ(k) = i, 1 i k. Show that τ ik σ leaves k fixed and hence is, in effect, a permutation of k Prove that if τ S k is a transposition, ( 1) τ = 1 and deduce from this Proposition

25 3. Prove assertion 2 in Proposition Prove that dim A k (V ) is given by (1.4.12). 5. Verify that for i < j 1 τ i,j = τ j 1,j τ i,j 1,τ j 1,j. 1.4 Alternating k-tensors For k = 3 show that every one of the six elements of S 3 is either a transposition or can be written as a product of two transpositions. 7. Let σ S k be the cyclic permutation σ(i) = i + 1, i = 1,...,k 1 and σ(k) = 1. Show explicitly how to write σ as a product of transpositions and compute ( 1) σ. Hint: Same hint as in exercise In exercise 7 of Section 3 show that if T is in A k, T v is in A k 1. Show in addition that for v,w V and T A k, (T v ) w = (T w ) v. 9. Let A : V W be a linear mapping. Show that if T is in A k (W), A T is in A k (V ). 10. In exercise 9 show that if T is in L k (W), Alt (A T) = A (Alt (T)), i.e., show that the Alt operation commutes with the pull-back operation.

26 26 Chapter 1. Multilinear algebra 1.5 The space, Λ k (V ) In 1.4 we showed that the image of the alternation operation, Alt : L k (V ) L k (V ) is A k (V ). In this section we will compute the kernel of Alt. Definition A decomposable k-tensor l 1 l k, l i V, is redundant if for some index, i, l i = l i+1. Let I k be the linear span of the set of reductant k-tensors. Note that for k = 1 the notion of redundant doesn t really make sense; a single vector l L 1 (V ) can t be redundant so we decree I 1 (V ) = {0}. Proposition If T I k, Alt (T) = 0. Proof. Let T = l k l k with l i = l i+1. Then if τ = τ i,i+1, T τ = T and ( 1) τ = 1. Hence Alt (T) = Alt (T τ ) = Alt (T) τ = Alt(T); so Alt (T) = 0. To simplify notation let s abbreviate L k (V ), A k (V ) and I k (V ) to L k, A k and I k. Proposition If T I r and T L s then T T and T T are in I r+s. Proof. We can assume that T and T are decomposable, i.e., T = l 1 l r and T = l 1 l s and that T is redundant: l i = l i+1. Then T T = l 1 l i 1 l i l i l r l 1 l s is redundant and hence in I r+s. The argument for T T is similar. Proposition If T L k and σ S k, then (1.5.1) T σ = ( 1) σ T + S where S is in I k.

27 1.5 The space, Λ k (V ) 27 Proof. We can assume T is decomposable, i.e., T = l 1 l k. Let s first look at the simplest possible case: k = 2 and σ = τ 1,2. Then T σ ( ) σ T = l 1 l 2 + l 2 l 1 = ((l 1 + l 2 ) (l 1 + l 2 ) l 1 l 1 l 2 l 2 )/2, and the terms on the right are redundant, and hence in I 2. Next let k be arbitrary and σ = τ i,i+1. If T 1 = l 1 l i 2 and T 2 = l i+2 l k. Then T ( 1) σ T = T 1 (l i l i+1 + l i+1 l i ) T 2 is in I k by Proposition and the computation above. The general case: By Theorem 1.4.2, σ can be written as a product of m elementary transpositions, and we ll prove (1.5.1) by induction on m. We ve just dealt with the case m = 1. The induction step: m 1 implies m. Let σ = τβ where β is a product of m 1 elementary transpositions and τ is an elementary transposition. Then T σ = (T β ) τ = ( 1) τ T β + = ( 1) τ ( 1) β T + = ( 1) σ T + where the dots are elements of I k, and the induction hypothesis was used in line 2. Corollary. If T L k, the (1.5.2) Alt (T) = k!t + W, where W is in I k. Proof. By definition Alt (T) = ( 1) σ T σ, and by Proposition 1.5.4, T σ = ( 1) σ T + W σ, with W σ I k. Thus Alt (T) = ( 1) σ ( 1) σ T + ( 1) σ W σ where W = ( 1) σ W σ. = k!t + W

28 28 Chapter 1. Multilinear algebra Corollary. I k is the kernel of Alt. Proof. We ve already proved that if T I k, Alt (T) = 0. To prove the converse assertion we note that if Alt (T) = 0, then by (1.5.2) with W I k. T = 1 k! W. Putting these results together we conclude: Theorem Every element, T, of L k can be written uniquely as a sum, T = T 1 + T 2 where T 1 A k and T 2 I k. Proof. By (1.5.2), T = T 1 + T 2 with and T 1 = 1 k! Alt (T) T 2 = 1 k! W. To prove that this decomposition is unique, suppose T 1 + T 2 = 0, with T 1 A k and T 2 I k. Then so T 1 = 0, and hence T 2 = 0. 0 = Alt (T 1 + T 2 ) = k!t 1 Let (1.5.3) Λ k (V ) = L k (V )/I k (V ), i.e., let Λ k = Λ k (V ) be the quotient of the vector space L k by the subspace, I k, of L k. By (1.2.3) one has a linear map: (1.5.4) π : L k Λ k, T T + I k which is onto and has I k as kernel. We claim: Theorem The map, π, maps A k bijectively onto Λ k. Proof. By Theorem every I k coset, T + I k, contains a unique element, T 1, of A k. Hence for every element of Λ k there is a unique element of A k which gets mapped onto it by π.

29 1.5 The space, Λ k (V ) 29 Remark. Since Λ k and A k are isomorphic as vector spaces many treatments of multilinear algebra avoid mentioning Λ k, reasoning that A k is a perfectly good substitute for it and that one should, if possible, not make two different definitions for what is essentially the same object. This is a justifiable point of view (and is the point of view taken by Spivak and Munkres 1 ). There are, however, some advantages to distinguishing between A k and Λ k, as we ll see in 1.6. Exercises. 1. A k-tensor, T, L k (V ) is symmetric if T σ = T for all σ S k. Show that the set, S k (V ), of symmetric k tensors is a vector subspace of L k (V ). 2. Let e 1,...,e n be a basis of V. Show that every symmetric 2- tensor is of the form aije i e j where a i,j = a j,i and e 1,...,e n are the dual basis vectors of V. 3. Show that if T is a symmetric k-tensor, then for k 2, T is in I k. Hint: Let σ be a transposition and deduce from the identity, T σ = T, that T has to be in the kernel of Alt. 4. Warning: In general S k (V ) I k (V ). Show, however, that if k = 2 these two spaces are equal. 5. Show that if l V and T I k 2, then l T l is in I k. 6. Show that if l 1 and l 2 are in V and T is in I k 2, then l 1 T l 2 + l 2 T l 1 is in I k. 7. Given a permutation σ S k and T I k, show that T σ I k. 8. Let W be a subspace of L k having the following two properties. (a) For S S 2 (V ) and T L k 2, S T is in W. (b) For T in W and σ S k, T σ is in W. 1 and by the author of these notes in his book with Alan Pollack, Differential Topology

30 30 Chapter 1. Multilinear algebra Show that W has to contain I k and conclude that I k is the smallest subspace of L k having properties a and b. 9. Show that there is a bijective linear map with the property α : Λ k A k (1.5.5) απ(t) = 1 Alt (T) k! for all T L k, and show that α is the inverse of the map of A k onto Λ k described in Theorem (Hint: 1.2, exercise 8). 10. Let V be an n-dimensional vector space. Compute the dimension of S k (V ). Some hints: (a) Introduce the following symmetrization operation on tensors T L k (V ): Sym(T) = T τ. τ S k Prove that this operation has properties 2, 3 and 4 of Proposition and, as a substitute for property 1, has the property: (SymT) σ = SymT. (b) Let ϕ I = Sym(e I ), e I = e i 1 e i n. Prove that {ϕ I, I non-decreasing} form a basis of S k (V ). (c) Conclude from (b) that dims k (V ) is equal to the number of non-decreasing multi-indices of length k: 1 i 1 i 2 l k n. (d) Compute this number by noticing that (i 1,...,i n ) (i 1 + 0,i 2 + 1,...,i k + k 1) is a bijection between the set of these non-decreasing multi-indices and the set of increasing multi-indices 1 j 1 < < j k n+k 1.

31 1.6 The wedge product The wedge product The tensor algebra operations on the spaces, L k (V ), which we discussed in Sections 1.2 and 1.3, i.e., the tensor product operation and the pull-back operation, give rise to similar operations on the spaces, Λ k. We will discuss in this section the analogue of the tensor product operation. As in 4 we ll abbreviate L k (V ) to L k and Λ k (V ) to Λ k when it s clear which V is intended. Given ω i Λ k i, i = 1,2 we can, by (1.5.4), find a T i L k i with ω i = π(t i ). Then T 1 T 2 L k 1+k 2. Let (1.6.1) ω 1 ω 2 = π(t 1 T 2 ) Λ k 1+k 2. Claim. This wedge product is well defined, i.e., doesn t depend on our choices of T 1 and T 2. Proof. Let π(t 1 ) = π(t 1 ) = ω 1. Then T 1 = T 1 + W 1 for some W 1 I k 1, so T 1 T 2 = T 1 T 2 + W 1 T 2. But W 1 I k 1 implies W 1 T 2 I k 1+k 2 and this implies: π(t 1 T 2 ) = π(t 1 T 2 ). A similar argument shows that (1.6.1) is well-defined independent of the choice of T 2. More generally let ω i Λ k i, i = 1,2,3, and let ω i = π(t i ), T i L k i. Define ω 1 ω 2 ω 3 Λ k 1+k 2 +k 3 by setting ω 1 ω 2 ω 3 = π(t 1 T 2 T 3 ). As above it s easy to see that this is well-defined independent of the choice of T 1, T 2 and T 3. It is also easy to see that this triple wedge product is just the wedge product of ω 1 ω 2 with ω 3 or, alternatively, the wedge product of ω 1 with ω 2 ω 3, i.e., (1.6.2) ω 1 ω 2 ω 3 = (ω 1 ω 2 ) ω 3 = ω 1 (ω 2 ω 3 ).

32 32 Chapter 1. Multilinear algebra We leave for you to check: For λ R (1.6.3) λ(ω 1 ω 2 ) = (λω 1 ) ω 2 = ω 1 (λω 2 ) and verify the two distributive laws: (1.6.4) and (1.6.5) (ω 1 + ω 2 ) ω 3 = ω 1 ω 3 + ω 2 ω 3 ω 1 (ω 2 + ω 3 ) = ω 1 ω 2 + ω 1 ω 3. As we noted in 1.4, I k = {0} for k = 1, i.e., there are no non-zero redundant k tensors in degree k = 1. Thus (1.6.6) Λ 1 (V ) = V = L 1 (V ). A particularly interesting example of a wedge product is the following. Let l i V = Λ 1 (V ), i = 1,...,k. Then if T = l 1 l k (1.6.7) l 1 l k = π(t) Λ k (V ). We will call (1.6.7) a decomposable element of Λ k (V ). We will prove that these elements satisfy the following wedge product identity. For σ S k : (1.6.8) l σ(1) l σ(k) = ( 1) σ l 1 l k. Proof. For every T L k, T = ( 1) σ T + W for some W I k by Proposition Therefore since π(w) = 0 (1.6.9) π(t σ ) = ( 1) σ π(t). In particular, if T = l 1 l k, T σ = l σ(1) l σ(k), so π(t σ ) = l σ(1) l σ(k) = ( 1) σ π(t) = ( 1) σ l 1 l k. In particular, for l 1 and l 2 V (1.6.10) l 1 l 2 = l 2 l 1

33 1.6 The wedge product 33 and for l 1, l 2 and l 3 V (1.6.11) l 1 l 2 l 3 = l 2 l 1 l 3 = l 2 l 3 l 1. More generally, it s easy to deduce from (1.6.8) the following result (which we ll leave as an exercise). Theorem If ω 1 Λ r and ω 2 Λ s then (1.6.12) ω 1 ω 2 = ( 1) rs ω 2 ω 1. Hint: It suffices to prove this for decomposable elements i.e., for ω 1 = l 1 l r and ω 2 = l 1 l s. Now make rs applications of (1.6.10). Let e 1,...,e n be a basis of V and let e 1,...,e n be the dual basis of V. For every multi-index, I, of length k, (1.6.13) e i 1 e i k = π(e I ) = π(e i 1 e i k ). Theorem The elements (1.6.13), with I strictly increasing, are basis vectors of Λ k. Proof. The elements ψ I = Alt (e I ), I strictly increasing, are basis vectors of A k by Proposition 3.6; so their images, π(ψ I ), are a basis of Λ k. But π(ψ I ) = π ( 1) σ (e I )σ = ( 1) σ π(e I )σ = ( 1) σ ( 1) σ π(e I) = k!π(e I ). Exercises: 1. Prove the assertions (1.6.3), (1.6.4) and (1.6.5). 2. Verify the multiplication law, (1.6.12) for wedge product.

34 34 Chapter 1. Multilinear algebra 3. Given ω Λ r let ω k be the k-fold wedge product of ω with itself, i.e., let ω 2 = ω ω, ω 3 = ω ω ω, etc. (a) Show that if r is odd then for k > 1, ω k = 0. (b) Show that if ω is decomposable, then for k > 1, ω k = If ω and µ are in Λ 2r prove: (ω + µ) k = k l=0 ( ) k ω l µ k l. l Hint: As in freshman calculus prove this binomial theorem by induction using the identity: ( ) ( k l = k 1 ) ( l 1 + k 1 ) l. 5. Let ω be an element of Λ 2. By definition the rank of ω is k if ω k 0 and ω k+1 = 0. Show that if ω = e 1 f e k f k with e i,f i V, then ω is of rank k. Hint: Show that ω k = k!e 1 f 1 e k f k. 6. Given e i V, i = 1,...,k show that e 1 e k 0 if and only if the e i s are linearly independent. Hint: Induction on k.

35 1.7 The interior product The interior product We ll describe in this section another basic product operation on the spaces, Λ k (V ). As above we ll begin by defining this operator on the L k (V ) s. Given T L k (V ) and v V let ι v T be the be the (k 1)-tensor which takes the value (1.7.1) k ι v T(v 1,...,v k 1 ) = ( 1) r 1 T(v 1,...,v r 1,v,v r,...,v k 1 ) r=1 on the k 1-tuple of vectors, v 1,...,v k 1, i.e., in the r th summand on the right, v gets inserted between v r 1 and v r. (In particular the first summand is T(v,v 1,...,v k 1 ) and the last summand is ( 1) k 1 T(v 1,...,v k 1,v).) It s clear from the definition that if v = v 1 + v 2 (1.7.2) ι v T = ι v1 T + ι v2 T, and if T = T 1 + T 2 (1.7.3) ι v T = ι v T 1 + ι v T 2, and we will leave for you to verify by inspection the following two lemmas: Lemma If T is the decomposable k-tensor l 1 l k then (1.7.4) ι v T = ( 1) r 1 l r (v)l 1 l r l k where the cap over l r means that it s deleted from the tensor product, and Lemma If T 1 L p and T 2 L q (1.7.5) ι v (T 1 T 2 ) = ι v T 1 T 2 + ( 1) p T 1 ι v T 2. We will next prove the important identity (1.7.6) ι v (ι v T) = 0. Proof. It suffices by linearity to prove this for decomposable tensors and since (1.7.6) is trivially true for T L 1, we can by induction

36 36 Chapter 1. Multilinear algebra assume (1.7.6) is true for decomposible tensors of degree k 1. Let l 1 l k be a decomposable tensor of degree k. Setting T = l 1 l k 1 and l = l k we have by (1.7.5). Hence ι v (l 1 l k ) = ι v (T l) = ι v T l + ( 1) k 1 l(v)t ι v (ι v (T l)) = ι v (ι v T) l + ( 1) k 2 l(v)ι v T +( 1) k 1 l(v)ι v T. But by induction the first summand on the right is zero and the two remaining summands cancel each other out. V From (1.7.6) we can deduce a slightly stronger result: For v 1,v 2 (1.7.7) ι v1 ι v2 = ι v2 ι v1. Proof. Let v = v 1 + v 2. Then ι v = ι v1 + ι v2 so 0 = ι v ι v = (ι v1 + ι v2 )(ι v1 + ι v2 ) = ι v1 ι v1 + ι v1 ι v2 + ι v2 ι v1 + ι v2 ι v2 = ι v1 ι v2 + ι v2 ι v1 since the first and last summands are zero by (1.7.6). We ll now show how to define the operation, ι v, on Λ k (V ). We ll first prove Lemma If T L k is redundant then so is ι v T. Proof. Let T = T 1 l l T 2 where l is in V, T 1 is in L p and T 2 is in L q. Then by (1.7.5) ι v T = ι v T 1 l l T 2 +( 1) p T 1 ι v (l l) T 2 +( 1) p+2 T 1 l l ι v T 2.

37 1.7 The interior product 37 However, the first and the third terms on the right are redundant and ι v (l l) = l(v)l l(v)l by (1.7.4). Now let π be the projection (1.5.4) of L k onto Λ k and for ω = π(t) Λ k define (1.7.8) ι v ω = π(ι v T). To show that this definition is legitimate we note that if ω = π(t 1 ) = π(t 2 ), then T 1 T 2 I k, so by Lemma ι v T 1 ι v T 2 I k 1 and hence π(ι v T 1 ) = π(ι v T 2 ). Therefore, (1.7.8) doesn t depend on the choice of T. By definition ι v is a linear mapping of Λ k (V ) into Λ k 1 (V ). We will call this the interior product operation. From the identities (1.7.2) (1.7.8) one gets, for v,v 1,v 2 V ω Λ k, ω 1 Λ p and ω 2 Λ 2 (1.7.9) (1.7.10) (1.7.11) and (1.7.12) ι (v1 +v 2 )ω = ι v1 ω + ι v2 ω ι v (ω 1 ω 2 ) = ι v ω 1 ω 2 + ( 1) p ω 1 ι v ω 2 ι v (ι v ω) = 0 ι v1 ι v2 ω = ι v2 ι v1 ω. Moreover if ω = l 1 l k is a decomposable element of Λ k one gets from (1.7.4) (1.7.13) ι v ω = k ( 1) r 1 l r (v)l 1 l r l k. r=1 In particular if e 1,...,e n is a basis of V, e 1,...,e n the dual basis of V and ω I = e i 1 e i k, 1 i 1 < < i k n, then ι(e j )ω I = 0 if j / I and if j = i r (1.7.14) ι(e j )ω I = ( 1) r 1 ω Ir where I r = (i 1,...,î r,...,i k ) (i.e., I r is obtained from the multiindex I by deleting i r ).

38 38 Chapter 1. Multilinear algebra Exercises: 1. Prove Lemma Prove Lemma Show that if T A k, i v = kt v where T v is the tensor (1.3.16). In particular conclude that i v T A k 1. (See 1.4, exercise 8.) 4. Assume the dimension of V is n and let Ω be a non-zero element of the one dimensional vector space Λ n. Show that the map (1.7.15) ρ : V Λ n 1, v ι v Ω, is a bijective linear map. Hint: One can assume Ω = e 1 e n where e 1,...,e n is a basis of V. Now use (1.7.14) to compute this map on basis elements. 5. (The cross-product.) Let V be a 3-dimensional vector space, B an inner product on V and Ω a non-zero element of Λ 3. Define a map by setting V V V (1.7.16) v 1 v 2 = ρ 1 (Lv 1 Lv 2 ) where ρ is the map (1.7.15) and L : V V the map (1.2.9). Show that this map is linear in v 1, with v 2 fixed and linear in v 2 with v 1 fixed, and show that v 1 v 2 = v 2 v For V = R 3 let e 1, e 2 and e 3 be the standard basis vectors and B the standard inner product. (See 1.1.) Show that if Ω = e 1 e 2 e 3 the cross-product above is the standard cross-product: (1.7.17) e 1 e 2 = e 3 e 2 e 3 = e 1 e 3 e 1 = e 2. Hint: If B is the standard inner product Le i = e i. Remark One can make this standard cross-product look even more standard by using the calculus notation: e 1 = î, e 2 = ĵ and e 3 = k

39 1.8 The pull-back operation on Λ k 1.8 The pull-back operation on Λ k 39 Let V and W be vector spaces and let A be a linear map of V into W. Given a k-tensor, T L k (W), the pull-back, A T, is the k-tensor (1.8.1) A T(v 1,...,v k ) = T(Av 1,...,Av k ) in L k (V ). (See 1.3, equation ) In this section we ll show how to define a similar pull-back operation on Λ k. Lemma If T I k (W), then A T I k (V ). Proof. It suffices to verify this when T is a redundant k-tensor, i.e., a tensor of the form T = l 1 l k where l r W and l i = l i+1 for some index, i. But by (1.3.14) A T = A l 1 A l k and the tensor on the right is redundant since A l i = A l i+1. Now let ω be an element of Λ k (W ) and let ω = π(t) where T is in L k (W). We define (1.8.2) A ω = π(a T). Claim: The left hand side of (1.8.2) is well-defined. Proof. If ω = π(t) = π(t ), then T = T + S for some S I k (W), and A T = A T + A S. But A S I k (V ), so Proposition The map π(a T ) = π(a T). A : Λ k (W ) Λ k (V ), mapping ω to A ω is linear. Moreover,

40 40 Chapter 1. Multilinear algebra (i) If ω i Λ k i (W), i = 1,2, then (1.8.3) A (ω 1 ω 2 ) = A ω 1 A ω 2. (ii) If U is a vector space and B : U V a linear map, then for ω Λ k (W ), (1.8.4) B A ω = (AB) ω. We ll leave the proof of these three assertions as exercises. Hint: They follow immediately from the analogous assertions for the pullback operation on tensors. (See (1.3.14) and (1.3.15).) As an application of the pull-back operation we ll show how to use it to define the notion of determinant for a linear mapping. Let V be a n-dimensional vector space. Then dim Λ n (V ) = ( n n) = 1; i.e., Λ n (V ) is a one-dimensional vector space. Thus if A : V V is a linear mapping, the induced pull-back mapping: A : Λ n (V ) Λ n (V ), is just multiplication by a constant. We denote this constant by det(a) and call it the determinant of A, Hence, by definition, (1.8.5) A ω = det(a)ω for all ω in Λ n (V ). From (1.8.5) it s easy to derive a number of basic facts about determinants. Proposition If A and B are linear mappings of V into V, then (1.8.6) det(ab) = det(a) det(b). Proof. By (1.8.4) and (AB) ω = det(ab)ω so, det(ab) = det(a) det(b). = B (A ω) = det(b)a ω = det(b) det(a)ω,

41 1.8 The pull-back operation on Λ k 41 Proposition If I : V V is the identity map, Iv = v for all v V, det(i) = 1. We ll leave the proof as an exercise. Hint: I is the identity map on Λ n (V ). Proposition If A : V V is not onto, det(a) = 0. Proof. Let W be the image of A. Then if A is not onto, the dimension of W is less than n, so Λ n (W ) = {0}. Now let A = I W B where I W is the inclusion map of W into V and B is the mapping, A, regarded as a mapping from V to W. Thus if ω is in Λ n (V ), then by (1.8.4) A ω = B I W ω and since I W ω is in Λn (W) it is zero. We will derive by wedge product arguments the familiar matrix formula for the determinant. Let V and W be n-dimensional vector spaces and let e 1,...,e n be a basis for V and f 1,...,f n a basis for W. From these bases we get dual bases, e 1,...,e n and f 1,...,f n, for V and W. Moreover, if A is a linear map of V into W and [a i,j ] the n n matrix describing A in terms of these bases, then the transpose map, A : W V, is described in terms of these dual bases by the n n transpose matrix, i.e., if then Ae j = a i,j f i, A f j = a j,i e i. (See 2.) Consider now A (f 1 f n). By (1.8.3) A (f1 fn) = A f1 A fn = (a 1,k1 e k 1 ) (a n,kn e k n ) the sum being over all k 1,...,k n, with 1 k r n. Thus, A (f1 fn) = a 1,k1...a n,kn e k 1 e k n.

42 42 Chapter 1. Multilinear algebra If the multi-index, k 1,...,k n, is repeating, then e k 1 e k n is zero, and if it s not repeating then we can write k i = σ(i) i = 1,...,n for some permutation, σ, and hence we can rewrite A (f 1 f n ) as the sum over σ S n of a1,σ(1) a n,σ(n) (e 1 e n )σ. But so we get finally the formula (e 1 e n )σ = ( 1) σ e 1 e n (1.8.7) A (f 1 f n) = det[a i,j ]e 1 e n where (1.8.8) det[a i,j ] = ( 1) σ a 1,σ(1) a n,σ(n) summed over σ S n. The sum on the right is (as most of you know) the determinant of [a i,j ]. Notice that if V = W and e i = f i, i = 1,...,n, then ω = e 1 e n = f 1 f n, hence by (1.8.5) and (1.8.7), (1.8.9) det(a) = det[a i,j ]. Exercises. 1. Verify the three assertions of Proposition Deduce from Proposition a well-known fact about determinants of n n matrices: If two columns are equal, the determinant is zero. 3. Deduce from Proposition another well-known fact about determinants of n n matrices: If one interchanges two columns, then one changes the sign of the determinant. Hint: Let e 1,...,e n be a basis of V and let B : V V be the linear mapping: Be i = e j, Be j = e i and Be l = e l, l i,j. What is B (e 1 e n )?

43 1.8 The pull-back operation on Λ k Deduce from Propositions and another well-known fact about determinants of n n matrix. If [b i,j ] is the inverse of [a i,j ], its determinant is the inverse of the determinant of [a i,j ]. 5. Extract from (1.8.8) a well-known formula for determinants of 2 2 matrices: [ ] a11, a det 12 = a a 21, a 11 a 22 a 12 a Show that if A = [a i,j ] is an n n matrix and A t = [a j,i ] is its transpose deta = deta t. Hint: You are required to show that the sums ( 1) σ a 1,σ(1)...a n,σ(n) σ S n and ( 1) σ a σ(1),1...a σ(n),n σ S n are the same. Show that the second sum is identical with ( 1) τ a τ(1),1...a τ(n),n summed over τ = σ 1 S n. 7. Let A be an n n matrix of the form [ ] B A = 0 C where B is a k k matrix and C the l l matrix and the bottom l k block is zero. Show that deta = det B detc. Hint: Show that in (1.8.8) every non-zero term is of the form where σ S k and τ S l. ( 1) στ b 1,σ(1)...b k,σ(k) c 1,τ(1)...c l,τ(l) 8. Let V and W be vector spaces and let A : V W be a linear map. Show that if Av = w then for ω Λ p (w ), A ι(w)ω = ι(v)a ω. (Hint: By (1.7.10) and proposition it suffices to prove this for ω Λ 1 (W ), i.e., for ω W.)